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If I'm not mistaken, $\mathbb{Q} / \mathbb{Z}$ are equivalence classes of rationals where $q \sim q^\prime$ iff they differ by an integer. So I could equally imagine this set as $\mathbb{Q} \cap [0,1]$ (mod 1) (if you know what I mean?).

Is that the right way to think about this group? I'm asking because I would like to see why elements in $\mathbb{Q} / \mathbb{Z}$ have finite order.

Many thanks for your help.

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I assume you are thinking of these as groups under addition. Then each equivalence class has a unique (rational) representative in $[0,1)$. Seems to me like a very good way of thinking concretely about the quotient group. – André Nicolas Oct 13 '11 at 16:29
Thank you! I'm just never sure whether what I do is right... – Rudy the Reindeer Oct 13 '11 at 16:35
You can also think of $\mathbb Q/\mathbb Z$ as the group of roots of unity in $\mathbb C$. – Pierre-Yves Gaillard Oct 13 '11 at 16:59
This question is related to Pierre's comment. – Martin Sleziak Aug 4 '12 at 11:18
@MartinSleziak Thanks. By now I finally managed to understand the two ($S^1$ and $\mathbb T$ or same for $\mathbb Q$). It's not clear to me anymore what I didn't understand. – Rudy the Reindeer Aug 4 '12 at 11:23
up vote 9 down vote accepted

Yes, $\mathbb{Q}\cap [0,1)$ forms a set of representatives for $\mathbb{Q}/\mathbb{Z}$, and "mod 1" is a good way of thinking about it. To see why an element $\left[\frac{a}{b}\right]\in\mathbb{Q}/\mathbb{Z}$ has finite order (I am using the brackets to mean the equivalence class of $\frac{a}{b}$), consider that this is equivalent to the statement that there is a finite number $n$ such that when I add $\frac{a}{b}$ to itself $n$ times, I get an integer. Do you see an $n$ that works?

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$b$ might work? – Rudy the Reindeer Oct 13 '11 at 16:37
@Matt: Precisely! :) – Zev Chonoles Oct 13 '11 at 16:45

Yes: the elements of $[0,1)\cap\mathbb{Q}$ form a complete set of coset representatives for $\mathbb{Z}$ in $\mathbb{Q}$. That is:

  1. Every element of $\mathbb{Q}$ is congruent modulo $\mathbb{Z}$ to at least one element of $[0,1)\cap\mathbb{Q}$; and
  2. No two distinct elements of $[0,1)\cap\mathbb{Q}$ are congruent modulo $\mathbb{Z}$.

That means that each element of $\mathbb{Q}/\mathbb{Z}$ corresponds to one and only one element of $[0,1)\cap \mathbb{Q}$. Since the sum of classes in the quotient is equal to the class of the sum, we have that if $q_1,q_2\in\mathbb{Q}$ correspond to the rationals $r_1$ and $r_2$ in $[0,1)\cap\mathbb{Q}$, respectively, then $$\begin{align*} (q_1+\mathbb{Z}) + (q_2+\mathbb{Z}) &= (r_1+\mathbb{Z}) + (r_2+\mathbb{Z})\\ &= (r_1+r_2)+\mathbb{Z}\\ &=\left\{\begin{array}{ll} (r_1+r_2)+\mathbb{Z} & \text{if }0\leq r_1+r_2\lt 1\\ (r_1+r_2-1 + \mathbb{Z} & \text{if }1\leq r_1+r_2 \end{array}\right. \end{align*}$$ which shows you that this group is isomorphic to $\mathbb{Q}\cap[0,1]$ mod $1$.

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If one is interested in the topology, then it is better to use the circle: $S \cap \mathbb{Q}$. – André Caldas Oct 13 '11 at 16:46
Um, the circle intersected with Q is {1,-1}. Also, Q/Z doesn't have an automatic topology. – KCd Oct 14 '11 at 2:53

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